Respuesta :
Answer:
Explanation:
Given
Radius of bicycle wheel [tex]r=0.3\ m[/tex]
Initial angular velocity [tex]\omega _0=0[/tex]
It rotates 3 revolution in 5 s therefore
[tex]\omega =2\pi 3=\6\pi =18.85\ rad/s[/tex]
using [tex]\omega =\omega _0+\alpha t[/tex]
where [tex]\alpha =angular\ acceleration[/tex]
[tex]\omega =Final\ angular\ velocity[/tex]
[tex]t=time[/tex]
[tex]\alpha =\frac{18.85}{5}=3.77 rad/s^2[/tex]
Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration
[tex]\omega at t=1[/tex]
[tex]\omega =0+3.77\times 1=3.77 rad/s[/tex]
[tex]a_c=\omega ^2\cdot r[/tex]
[tex]a_c=(3.77)^2\cdot 0.3=4.26 m/s^2[/tex]
Tangential acceleration [tex]a_t=\alpha \times r[/tex]
[tex]a_t=3.77\times 0.3=1.13 m/s^2[/tex]
[tex]a_{net}=\sqrt{a_t^2+a_c^2}[/tex]
[tex]a_{net}=\sqrt{(1.13)^2+(4.26)^2}[/tex]
[tex]a_{net}=4.41 m/s^2[/tex]
The magnitude of the acceleration is 4.47m/s^2 and the direction is 75.14° towards the edge of the wheel
Data;
- radius = 0.30m
- angular acceleration = 3 rev/s
- time = 5s
Magnitude and Direction of Total Acceleration of Vector
Radius (R)=0.3 m
Initial angular velocity = [tex](\omega_1)= 0 rad/s[/tex]
Final angular velocity [tex]\omega _f = 3rev/s[/tex]
[tex]\omega_f= (3*2\pi ) rad/s =6\pi rad/s[/tex]
Using equation of angular acceleration
[tex]\omega_f= \omega_i +\alpha*t[/tex]
[tex]\alpha= (6\pi /5)= 1.2\pi rad/s2at= R\alpha= (0.3*1.2\pi )= 1.131 m/s^2 (towards tangent)a_c= \omega ^2R = (\alpha t)^2*Ra_c= (1.2\pi *1)2*0.3a_c= (1.44\pi *\pi *0.3)= 4.2636 m/s^2 (towards center)\\a= \sqrt{(at^2+ac^2)} \\a= \sqrt{(1.131)^2+(4.2636)^2} a=\sqrt{(20.058}[/tex]
The acceleration (magnitude)= 4.4786 m/s2
The direction can be calculated as
[tex]\theta= tan^-^1(a_c/a_t)[/tex]
Substituting the values,
[tex]\theta = tan^-^1(4.2636/1.131)\\\theta = 75.14^0[/tex]
The magnitude of the acceleration is 4.47m/s^2 and the direction is 75.14° towards the edge of the wheel
Learn more on centripetal acceleration here;
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